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Sergeeva-Olga [200]
3 years ago
13

Children’s chewable tylenol contains 80 mg of acetaminophen per tablet. if the recommended dosage is 10 mg/kg, how many tablets

are needed for a 53−lb child?
Chemistry
2 answers:
vovikov84 [41]3 years ago
8 0

Answer:

3 tablets are needed for a 53-lb child

Explanation:

1 lb = 0.4536 kg

So, 53 lb = (53\times 0.4536)kg=24 kg

So, weight of the child is 24 kg

For 1 kg body, recommended dosage is 10 mg

Hence, for 24 kg body, recommended dosage is (10\times 24)mg or 240 mg

80 mg of acetaminophen is present in 1 tablet

So, 240 mg acetaminophen is present in \frac{1}{80}\times 240 tablets or 3 tablets.

So, 3 tablets are needed for a 53-lb child

eduard3 years ago
6 0

53 pounds divided by 2.2 kilograms to get the converted weight which is 24.09 kilograms. Round this off to 24 kgs. 10 mg/kg multiply this to 24 kg, it would be 240 mg. 240 mg divided by 80 is 3. Therefore, a 53−lb child requires to have a dosage of 3 tablets.

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How many moles of FeCI3 are present in 345.0 g FeCI3
Sphinxa [80]

Answer:

\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

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There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

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2. Calculate Moles

Use the number as a ratio.

\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Multiply by the given number of grams, 345.0.

345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Flip the fraction so the grams of FeCl₃ will cancel.

345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}

345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }

\frac{345.0 \ mol \ FeCl_3}{162.19 }

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2.12713484 \ mol \ FeCl_3

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

\approx 2.127 \ mol \ FeCl_3

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

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